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If $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ are in harmonical progression such that $\mathrm{a}>\mathrm{d}$, then which one of the following is correct?
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The correct answer is:
$a+c>b+d$
Since, $\mathrm{a}>\mathrm{d}$ and $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ are in $\mathrm{HP}$.
$\Rightarrow \mathrm{a}>\mathrm{b}>\mathrm{c}>\mathrm{d}$.
$\mathrm{b}=\frac{2 \mathrm{a} \mathrm{c}}{\mathrm{a}+\mathrm{c}} \Rightarrow \mathrm{b}^{2} \mathrm{~d}=\frac{2 \mathrm{a} \mathrm{b} \mathrm{cd}}{\mathrm{a}+\mathrm{c}}$
and $c=\frac{2 b d}{b+d} \Rightarrow c^{2} a=\frac{2 a b c d}{b+d}$
$\frac{c^{2} a}{b^{2} d}=\frac{a+c}{b+d}$
$\Rightarrow \quad \frac{a+c}{b+d}=\left(\frac{a}{b}\right) \cdot\left(\frac{c}{d}\right) \cdot\left(\frac{c}{b}\right)>1$
$\Rightarrow a+c>b+d$
$\Rightarrow \mathrm{a}>\mathrm{b}>\mathrm{c}>\mathrm{d}$.
$\mathrm{b}=\frac{2 \mathrm{a} \mathrm{c}}{\mathrm{a}+\mathrm{c}} \Rightarrow \mathrm{b}^{2} \mathrm{~d}=\frac{2 \mathrm{a} \mathrm{b} \mathrm{cd}}{\mathrm{a}+\mathrm{c}}$
and $c=\frac{2 b d}{b+d} \Rightarrow c^{2} a=\frac{2 a b c d}{b+d}$
$\frac{c^{2} a}{b^{2} d}=\frac{a+c}{b+d}$
$\Rightarrow \quad \frac{a+c}{b+d}=\left(\frac{a}{b}\right) \cdot\left(\frac{c}{d}\right) \cdot\left(\frac{c}{b}\right)>1$
$\Rightarrow a+c>b+d$
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